> That is true in the list of all FIS too> d_1> d_1, d_2> d_1, d_2, d_3> ...> > All these FIS however, are not capable of distinguishing themselves from all rationals. That requires more: A good portion of faith. And a completed infinity for that np larger index is possible. Obviously the largest index has been applied. An index which is not present in any rational number.>

We need only show that for each Rational Number, r, there is One particular FIS(d_n) such that we have r ~= FIS(d_n). This is a Sufficient Condition to conclude that r =~ FIS(d_n).

> He did not avoid it. He knew fairly early, in the 1890s, that the set of all sets is impossible. Unfortunately he did not recognize that the set of all natural numbers that are followed by infinitely many natural numbers is impossible too. (I.e., the set of all numbers that is not all numbers.)

He didn't realize that, because it is not true.

How does N = { All Natural Numbers } --> N = { All Numbers that is not All Numbers }?You exposition is bizarre.

What we can prove, in Modern Mathematics, is that the Union of the Set of All Finite Ordinals is Not a Finite Ordinal and, also, that the Cardinality of the Set of All Finite Cardinals is Not Finite.

> The usual reply of matheologians, on the question how they know that they have all numbers without having the last one, is: There is no last natural number. But I im interested to see their escape with respect to the fact that no digit d_i ad no FIS is sufficient to accomplish a distinction of d from all rationals of the rationals-complete list. Obviously d contains somewhat more than every digit d_i.>

We "have them all" because there is a Set that contains "them all".

The "escape" is accomplished by knowing that in the Set of All FIS of d, there is (at least) one k, such that FIS(d_k) ~= r, where is an Arbitrarily Chosen Rational Number.